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strojnjashka [21]
3 years ago
13

A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)

What is the player's speed afterward if the ball is thrown at 15 m/s relative to the ground?B) What is the player's speed afterward if the ball is thrown at 15 m/s relative to the player?
Physics
1 answer:
lord [1]3 years ago
4 0

Answer:

4.53482 m/s

4.506 m/s

Explanation:

m_1 = Mass of player = 75 kg

v_1 = Initial velocity of player = 4.6 m/s

m_2 = Mass of ball = 0.47 kg

v_1 = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s

The player's speed is 4.506 m/s

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Face width (F) = 1 in

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Diameter (d) = \frac{N}{P}

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To find the velocity (V); we use the formula:

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K_v = \frac{2000+V}{2000}

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However, there is need to get the value of the tangential load(W^t), in order to achieve that, we have the following expression

W^t=\frac{T}{\frac{d}{2} }

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W^t = 472.69 lbf

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\sigma = \frac{K_vW^tp}{FY}

\sigma = \frac{2.0472*472.69*5}{1*0.321}

\sigma = 15073.07 psi

\sigma = 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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